Leetcode:20. Valid Parentheses(JAVA)
2016-03-14 15:46
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【问题描述】
Given a string containing just the characters
determine if the input string is valid.
The brackets must close in the correct order,
all valid but
not.
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【思路】
1、使用栈检查符号是否成pair。
2、为了书写方便,将三对character放在map中使用较为方便。
【code】
public class Solution {
public boolean isValid(String s) {
HashMap<Character, Character> map = new HashMap<Character, Character>();
map.put('(', ')');
map.put('[', ']');
map.put('{', '}');
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char curr = s.charAt(i);
if (map.keySet().contains(curr)) {
stack.push(curr);
} else if (map.values().contains(curr)) {
if (!stack.isEmpty() && map.get(stack.peek()) == curr) {
stack.pop();
} else {
return false;
}
}
}
return stack.isEmpty();
}
}
Given a string containing just the characters
'(',
')',
'{',
'}',
'['and
']',
determine if the input string is valid.
The brackets must close in the correct order,
"()"and
"()[]{}"are
all valid but
"(]"and
"([)]"are
not.
Subscribe to see which companies asked this question
【思路】
1、使用栈检查符号是否成pair。
2、为了书写方便,将三对character放在map中使用较为方便。
【code】
public class Solution {
public boolean isValid(String s) {
HashMap<Character, Character> map = new HashMap<Character, Character>();
map.put('(', ')');
map.put('[', ']');
map.put('{', '}');
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char curr = s.charAt(i);
if (map.keySet().contains(curr)) {
stack.push(curr);
} else if (map.values().contains(curr)) {
if (!stack.isEmpty() && map.get(stack.peek()) == curr) {
stack.pop();
} else {
return false;
}
}
}
return stack.isEmpty();
}
}
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